Question Number 133536 by mathmax by abdo last updated on 22/Feb/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \mathrm{with}\:\mathrm{verify}\:\mathrm{u}_{\mathrm{n}} +\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\:\sqrt{\mathrm{n}}}\:\:\forall\mathrm{n}\geqslant\mathrm{1} \\ $$$$\mathrm{determine}\:\mathrm{a}\:\mathrm{equivalent}\:\mathrm{of}\:\:\mathrm{u}_{\mathrm{n}} \\ $$$$\mathrm{is}\:\mathrm{u}_{\mathrm{n}} \mathrm{convergent}? \\ $$ Terms of…
Question Number 2466 by prakash jain last updated on 20/Nov/15 $$\sqrt{−\frac{\mathrm{1}}{\mathrm{5}}\:}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=? \\ $$ Answered by Yozzi last updated on 20/Nov/15 $$\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}={i}\sqrt{\frac{\mathrm{1}}{\mathrm{5}}}−\frac{\mathrm{1}}{{i}\sqrt{\mathrm{5}}} \\ $$$$={i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}+\frac{{i}}{\:\sqrt{\mathrm{5}}}=\mathrm{2}\frac{{i}}{\:\sqrt{\mathrm{5}}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}=\frac{\sqrt{\mathrm{1}}}{\:\sqrt{−\mathrm{5}}}=\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}\Rightarrow\sqrt{\frac{\mathrm{1}}{−\mathrm{5}}}−\frac{\mathrm{1}}{\:\sqrt{−\mathrm{5}}}\overset{?}…
Question Number 133538 by mathmax by abdo last updated on 22/Feb/21 $$\mathrm{calculate}\:\int\int_{\left[\mathrm{1},\mathrm{2}\right]^{\mathrm{2}} } \:\:\:\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{3y}^{\mathrm{2}} }\mathrm{e}^{−\left(\mathrm{x}^{\mathrm{2}} \:+\mathrm{3y}^{\mathrm{2}} \right)} \mathrm{dxdy} \\ $$ Terms of Service Privacy…
Question Number 68001 by mathmax by abdo last updated on 03/Sep/19 $${let}\:{F}\left({x}\right)=\int_{\mathrm{2}{x}} ^{{x}^{\mathrm{2}} +\mathrm{1}} \:\:\frac{{e}^{−{xt}} }{{x}+\mathrm{2}{t}}{dt}\:\:\:\:{calculate}\:{F}\:^{'} \left({x}\right) \\ $$ Commented by mathmax by abdo last…
Question Number 133533 by Raxreedoroid last updated on 22/Feb/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{that}\:\mathrm{is}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curves} \\ $$$${y}={x}^{\mathrm{3}} ,{y}=\mathrm{8},{x}=\mathrm{0},\:\mathrm{rotated}\:\mathrm{about}\:{x}=\mathrm{9} \\ $$ Answered by bemath last updated on 23/Feb/21 $$\mathrm{V}=\pi\underset{\mathrm{0}} {\overset{\mathrm{8}} {\int}}\left(\mathrm{9}−\sqrt[{\mathrm{3}}]{\mathrm{y}}\:\right)^{\mathrm{2}}…
Question Number 133535 by mathmax by abdo last updated on 22/Feb/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{sequence}\:\mathrm{u}_{\mathrm{n}} \mathrm{wich}\:\mathrm{verify}\:\:\mathrm{u}_{\mathrm{n}} +\mathrm{u}_{\mathrm{n}+\mathrm{1}} =\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:\forall\mathrm{n}\geqslant\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 67996 by TawaTawa last updated on 03/Sep/19 Commented by mathmax by abdo last updated on 03/Sep/19 $${we}\:{have}\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{n}^{\mathrm{2}} +\mathrm{1}\leqslant{n}^{\mathrm{2}} \:+{k}\leqslant{n}^{\mathrm{2}} \:+{n}\:\Rightarrow \\ $$$$\sqrt{{n}^{\mathrm{2}} \:+\mathrm{1}}\leqslant\sqrt{{n}^{\mathrm{2}}…
Question Number 67997 by ajfour last updated on 03/Sep/19 $$\mathrm{5}{y}^{\mathrm{2}} +\mathrm{2}{axy}+{b}=\mathrm{0} \\ $$$${ay}^{\mathrm{2}} +\mathrm{2}{bx}+\mathrm{5}{c}=\mathrm{0} \\ $$$$\left(\mathrm{5}{x}+\mathrm{3}{a}\right){y}^{\mathrm{2}} +\left(\mathrm{4}{ax}^{\mathrm{2}} \right){y}−{bx}−\mathrm{5}{c}=\mathrm{0} \\ $$$$\mathrm{5}{y}^{\mathrm{2}} −{x}\left(\mathrm{5}{x}+\mathrm{2}{a}\right){y}−{ax}^{\mathrm{3}} −\mathrm{3}{b}=\mathrm{0} \\ $$$${Please}\:{solve}\:{simultaneously} \\…
Question Number 2458 by alib last updated on 20/Nov/15 $${The}\:{medians}\:{of}\:{a}\:{triangle} \\ $$$${are}\:{m}_{\mathrm{1}} ,\:{m}_{\mathrm{2}} ,\:{m}_{\mathrm{3}} . \\ $$$${Find}\:{the}\:{length}\:{of}\:{each}\:{sides}\: \\ $$$${the}\:{triangle}. \\ $$ Answered by prakash jain…
Question Number 67992 by MJS last updated on 03/Sep/19 $$\left(\mathrm{1}\right)\:{z}={a}+{b}\mathrm{i} \\ $$$$\left(\mathrm{2}\right)\:{z}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\mathrm{express}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{real}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{real}\:\mathrm{part}\right] \\ $$$$\left(\mathrm{b}\right)\:\mathrm{imag}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{imaginary}\:\mathrm{part}\right] \\ $$$$\left(\mathrm{c}\right)\:\mathrm{abs}\:\left({z}^{{z}} \right)\:\:\:\:\:\left[\mathrm{absolute}\:\mathrm{value}\right] \\ $$$$\left(\mathrm{d}\right)\:\mathrm{arg}\:\left({z}^{{z}}…